Difficulty solving &#x222C;<!-- ∬ --> ln &#x2061;<!-- ⁡ --> ( x + y 2 </ms

Sovardipk 2022-07-03 Answered
Difficulty solving ln ( x + y 2 ) y 2 d x d y with change of variables
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Pranav Greer
Answered 2022-07-04 Author has 13 answers
Let me try to guess. You are interested in a function f:
2 x y f ( x , y ) = y 2 ln ( x + y 2 ) .
You attempt to do a substitution u = x + y 2 and find function g ( u ( x , y ) , v ( y ) ) = f ( x , y )
y ( x g ( u , v ) ) = y ( g ( u , v ) u u x ) = y ( g ( u , v ) u ) = 2 g u 2 u y + 2 g u v v y = ( 2 y 2 u 2 + 2 y 2 u v ) g = 2 v ( 2 u 2 + 2 u v ) g .
So we see that operator 2 x y doesn't transform into ( u , v ) ( x , y ) 2 u v the same way as dxdy transforms to ( x , y ) ( u , v ) d u d v. Whereas in 1D q is indeed p q p . This is the reason why you can treat derivatives as fractions of differentials in 1D, but cannot do the same in 2D+.
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more